Check if record does NOT exist in Rails (from array of ids)?

Question

I can do this to check if a record(s) exists (say id "1" exists, but "2" and "3" don't):

Model.exists?(:id => [1, 2, 3]) #=> true

How do I do the opposite, so:

Model.not_exists?(:id => [1, 2, 3]) #=> true

Bluecollar

Answer 1

Edit: Please don't use this method. Even if it works it's not optimal since it loads all records instead of only testing their existence. This is a better way of doing it.

---

If you only need search records through ID you can try this

class Model
  def self.not_exists?(ids)
    self.find(ids)
    false
  rescue
    true
  end
end

If any of the IDs does not exist the find method will raise a ActiveRecord::RecordNotFound exception that we simply catch and return true.

Marilou

This should not be the accepted answer, rescue as its used is dangerous and find will load the records where you only care at the existence of a raw. Answer belong should be the one accepted – Barnstorm

Answer 2

just add a ! operator

!Model.exists?(:id => [1, 2, 3]) #=> true

Pommel

I'm asking how to say if 2 and 3 don't exist, how to make it return false. !Model.exists?(:id => [1, 2, 3]) returns false if 1 exists, whereas I want Model.not_exists?(:id => [1, 2, 3]) to return true if any don't exist. – Bluecollar

Model.find(ids_ary).count then rescue ActiveRecord::RecordNotFound – Hydrostat

Answer 3

Use empty?, this is what you want. It uses count(*) vs select 1 as one.

> Rocketeer.where(:id => [1, 2, 3]).empty?
   (0.6ms)  SELECT COUNT(*) FROM "rocketeers" WHERE "rocketeers"."id" IN (1, 2, 3)
=> false

> Rocketeer.where(:id => [1, 2, 3]).any?
   (0.5ms)  SELECT COUNT(*) FROM "rocketeers" WHERE "rocketeers"."id" IN (1, 2, 3)
=> true

> Rocketeer.where(:id => [1, 2, 3]).exists?
  Rocketeer Exists (0.5ms)  SELECT  1 AS one FROM "rocketeers" WHERE "rocketeers"."id" IN (1, 2, 3) LIMIT 1
=> true

Brufsky

Isn't SELECT 1 AS one ... LIMIT 1 more efficient than SELECT COUNT(*)? – Riebling

limit 1 is equal or faster than count(*). Although generally it'll be only perceptible with a combination of table size and index keys. Also, favor use of empty? to save DB RTT, if you'll use matching records. – Pye

Answer 4

A more Ruby-esque way of doing it would be to use unless with exists?. This way, you don't have to use !. I imagine your use case is something like this:

def my_method
    return unless Model.exists?(:id => [1, 2, 3])

    # do something
end

You can replace 1, 2, 3 with a variable (call it id or something) and even remove the array altogether if you'd like: .exists?(id: id)

Mangrum

Answer 5

class Model
  def self.does_not_exist?(ids)
    Model.where(id: ids).count < ids.size
  end
end

Explanation: If (and only if) all the instances you're looking for exist, Model.where(id: ids).count is equal to ids.size.

However if there are one or more instances missing, the count will be lower, meaning that there's a record that does not exist.

Golf

Maybe with an explanation it's more helpful – Upstart

It answers the question but is slower performance for a large number of rows. It does a COUNT in SQL which counts every row in a table. where(id: [1, 2, 3]).exists? on the other hand does a SELECT... WHERE... LIMIT 1 which will return first row it finds. See @Brufsky 's answer. – Binni

Answer 6

Another simple way is to use the where method with an array of id's.

# If the count of the query is equal to the count of all of the id's then the statement will return false.
# Else it will return true if not all ids exists in the database.
Model.where(id: [1, 2, 3]).count < [1,2,3].count

Ribbentrop

Answer 7

To ensure all ids exist, not just one the following works without have to rescue

ids = [1,2,3]
Model.where(id: ids).count != ids.length

The exists? method will return true if any of the ids match. The above will ensure all ids exist.

Charlsiecharlton

Answer 8

.none? returns true if there are no records.

Model.none?(id: [1, 2, 3])

Tannen